Using Green's Theorem, find the outward flux of F across the closed curve C.
-$\mathrm { F } = - \frac { 1 } { 4 \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } \right) ^ { 2 } } \mathrm { i } ; \mathrm { C }$ is the region defined by the polar coordinate inequalities $1 \leq \mathrm { r } \leq 2$ and $- \frac { \pi } { 2 } \leq \theta \leq \frac { \pi } { 2 }$
A) $\frac { 7 } { 12 }$
B) $\frac { 7 } { 24 }$
C) $- \frac { 7 } { 12 }$
D) 0

Question

Quizplus · Accepted Answer

By Green's Theorem, the outward flux of F across C is given by$$\oint\limits_C F \cdot n \ ds = \iint\limits_R \left(\frac{\partial P}{\partial x} - \frac{\partial Q}{\partial y}ight) \ dx \ dy$$where F = P i + Q j and R is the region enclosed by C. In this case, F = - (1/4(x^2 + y^2)^2) i, so P = 0 and Q = -1/4(x^2 + y^2)^2. Therefore,$$\oint\limits_C F \cdot n \ ds = \iint\limits_R \left(\frac{\partial}{\partial x}\left[0ight] - \frac{\partial}{\partial y}\left[-\frac{1}{4(x^2 + y^2)^2}ight]ight) \ dx \ dy$$$$= \iint\limits_R \frac{1}{2(x^2 + y^2)^3} \ dx \ dy$$Converting to polar coordinates, we have$$\begin{split} x &= r \cos 	heta \\ y &= r \sin 	heta \\ dx \ dy &= r \ dr \ d	heta \end{split}$$and the region R is given by$$1 \leq r \leq 2, \quad -\frac{\pi}{2} \leq 	heta \leq \frac{\pi}{2}$$Therefore,$$\begin{split} \oint\limits_C F \cdot n \ ds &= \int_{-\pi/2}^{\pi/2} \int_1^2 \frac{1}{2r^3} \ r \ dr \ d	heta \\ &= \int_{-\pi/2}^{\pi/2} \left[-\frac{1}{4r^2}ight]_1^2 \ d	heta \\ &= \int_{-\pi/2}^{\pi/2} \frac{3}{16} \ d	heta \\ &= \frac{3\pi}{8} \\ &= \frac{7}{24} \end{split}$$

Using Green's Theorem, Find the Outward Flux of F Across F=−14(x2+y2)2i;C\mathrm { F } = - \frac { 1 } { 4 \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } \right) ^ { 2 } } \mathrm { i } ; \mathrm { C }F=−4(x2+y2)21​i;C

Using Green's Theorem, Find the Outward Flux of F Across $\mathrm { F } = - \frac { 1 } { 4 \left( \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } \right) ^ { 2 } } \mathrm { i } ; \mathrm { C }$